A) \[2\]
B) \[2\sqrt{3}\]
C) \[\frac{1}{\sqrt{3}}\]
D) \[4\]
E) \[0\]
Correct Answer: A
Solution :
Given that, \[PS+PS=8\] \[PS=8-PS\] Squaring on both sides, \[{{(PS)}^{2}}=64+{{(PS)}^{2}}-16(PS)\] \[\{{{(x-2)}^{2}}+{{y}^{2}}\}=64+\{{{(x+2)}^{2}}+{{y}^{2}}\}\] \[-16\sqrt{{{(x-2)}^{2}}+{{y}^{2}}}\] \[[\because PS=\sqrt{{{(x+2)}^{2}}+{{y}^{2}}}]\] \[[\because PS=\sqrt{{{(x-2)}^{2}}+{{y}^{2}}}]\] \[({{x}^{2}}+{{y}^{2}}+4-4x)=({{x}^{2}}+{{y}^{2}}+4+4x+64)\] \[-16\sqrt{({{x}^{2}}+{{y}^{2}}+4x+4)}\] \[\Rightarrow \] \[16\sqrt{({{x}^{2}}+{{y}^{2}}+4+4x)}=8x+64\] \[\Rightarrow \] \[2\sqrt{{{x}^{2}}+{{y}^{2}}+4x+4}=(x+8)\] Squaring on both sides; \[4{{x}^{2}}+4{{y}^{2}}+16x+16={{x}^{2}}+64+16x\] \[\Rightarrow \]\[3{{x}^{2}}+4{{y}^{2}}=48,\]which is equation of an ellipse given\[(y=3),\because (x,3)\]lies on an ellipse. \[\Rightarrow \] \[3{{x}^{2}}+4\times 9=48\] \[\Rightarrow \] \[3{{x}^{2}}=12\] \[\Rightarrow \] \[{{x}^{2}}=4\] (here we take only positive value of\[x\].) \[\Rightarrow \] \[x=\pm 2\] \[\Rightarrow \] \[x=2\]You need to login to perform this action.
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